diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -1,47 +1,48 @@
-\chapter{Finite-Dimensional Irreducible Representations}
+\chapter{Finite-Dimensional Simple Modules}
-In this chapter we classify the finite-dimensional irreducible representations
-of a finite-dimensional semisimple Lie algebra \(\mathfrak{g}\) over \(K\). At
-the heart of our analysis of \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\)
-was the decision to consider the eigenspace decomposition
+In this chapter we classify the finite-dimensional simple
+\(\mathfrak{g}\)-modules for a finite-dimensional semisimple Lie algebra
+\(\mathfrak{g}\) over \(K\). At the heart of our analysis of
+\(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\) was the decision to consider
+the eigenspace decomposition
\begin{equation}\label{sym-diag}
- V = \bigoplus_\lambda V_\lambda
+ M = \bigoplus_\lambda M_\lambda
\end{equation}
This was simple enough to do in the case of \(\mathfrak{sl}_2(K)\), but the
rational behind it and the reason why equation (\ref{sym-diag}) holds are
harder to explain in the case of \(\mathfrak{sl}_3(K)\). The eigenspace
-decomposition associated with an operator \(V \to V\) is a very well-known
+decomposition associated with an operator \(M \to M\) is a very well-known
tool, and readers familiarized with basic concepts of linear algebra should be
used to this type of argument. On the other hand, the eigenspace decomposition
-of \(V\) with respect to the action of an arbitrary subalgebra \(\mathfrak{h}
-\subset \mathfrak{gl}(V)\) is neither well-known nor does it hold in general:
+of \(M\) with respect to the action of an arbitrary subalgebra \(\mathfrak{h}
+\subset \mathfrak{gl}(M)\) is neither well-known nor does it hold in general:
as previously stated, it may very well be that
\[
- \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda \subsetneq V
+ \bigoplus_{\lambda \in \mathfrak{h}^*} M_\lambda \subsetneq M
\]
We should note, however, that these two cases are not as different as they may
sound at first glance. Specifically, we can regard the eigenspace decomposition
-of a representation \(V\) of \(\mathfrak{sl}_2(K)\) with respect to the
-eigenvalues of the action of \(h\) as the eigenvalue decomposition of \(V\)
-with respect to the action of the subalgebra \(\mathfrak{h} = K h \subset
-\mathfrak{sl}_2(K)\). Furthermore, in both cases \(\mathfrak{h} \subset
-\mathfrak{sl}_n(K)\) is the subalgebra of diagonal matrices, which is Abelian.
-The fundamental difference between these two cases is thus the fact that \(\dim
-\mathfrak{h} = 1\) for \(\mathfrak{h} \subset \mathfrak{sl}_2(K)\) while \(\dim
-\mathfrak{h} > 1\) for \(\mathfrak{h} \subset \mathfrak{sl}_3(K)\). The
-question then is: why did we choose \(\mathfrak{h}\) with \(\dim \mathfrak{h} >
-1\) for \(\mathfrak{sl}_3(K)\)?
+of a \(\mathfrak{sl}_2(K)\)-module \(M\) with respect to the eigenvalues of the
+action of \(h\) as the eigenvalue decomposition of \(M\) with respect to the
+action of the subalgebra \(\mathfrak{h} = K h \subset \mathfrak{sl}_2(K)\).
+Furthermore, in both cases \(\mathfrak{h} \subset \mathfrak{sl}_n(K)\) is the
+subalgebra of diagonal matrices, which is Abelian. The fundamental difference
+between these two cases is thus the fact that \(\dim \mathfrak{h} = 1\) for
+\(\mathfrak{h} \subset \mathfrak{sl}_2(K)\) while \(\dim \mathfrak{h} > 1\) for
+\(\mathfrak{h} \subset \mathfrak{sl}_3(K)\). The question then is: why did we
+choose \(\mathfrak{h}\) with \(\dim \mathfrak{h} > 1\) for
+\(\mathfrak{sl}_3(K)\)?
The rational behind fixing an Abelian subalgebra \(\mathfrak{h}\) is a simple
one: we have seen in the previous chapter that representations of Abelian
algebras are generally much simpler to understand than the general case. Thus
-it make sense to decompose a given representation \(V\) of \(\mathfrak{g}\)
-into subspaces invariant under the action of \(\mathfrak{h}\), and then analyze
-how the remaining elements of \(\mathfrak{g}\) act on these subspaces. The
-bigger \(\mathfrak{h}\) is, the simpler our problem gets, because there are
-fewer elements outside of \(\mathfrak{h}\) left to analyze.
+it make sense to decompose a given \(\mathfrak{g}\)-module \(M\) of into
+subspaces invariant under the action of \(\mathfrak{h}\), and then analyze how
+the remaining elements of \(\mathfrak{g}\) act on these subspaces. The bigger
+\(\mathfrak{h}\) is, the simpler our problem gets, because there are fewer
+elements outside of \(\mathfrak{h}\) left to analyze.
Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h}
\subset \mathfrak{g}\), which leads us to the following definition.
@@ -63,8 +64,8 @@ Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h}
\begin{proof}
Notice that \(0 \subset \mathfrak{g}\) is an Abelian subalgebra whose
- elements act as diagonal operators via the adjoint representation. Indeed,
- \(0\), the only element of \(0 \subset \mathfrak{g}\), is such that
+ elements act as diagonal operators via the adjoint \(\mathfrak{g}\)-module.
+ Indeed, \(0\), the only element of \(0 \subset \mathfrak{g}\), is such that
\(\operatorname{ad}(0) = 0\) is a diagonalizable operator. Furthermore, given
a chain of Abelian subalgebras
\[
@@ -101,12 +102,12 @@ The intersection of such subalgebra with \(\mathfrak{sl}_n(K)\) -- i.e. the
subalgebra of traceless diagonal matrices -- is a Cartan subalgebra of
\(\mathfrak{sl}_n(K)\). In particular, if \(n = 2\) or \(n = 3\) we get to the
subalgebras described the previous chapter. The remaining question then is: if
-\(\mathfrak{h} \subset \mathfrak{g}\) is a Cartan subalgebra and \(V\) is a
-representation of \(\mathfrak{g}\), does the eigenspace decomposition
+\(\mathfrak{h} \subset \mathfrak{g}\) is a Cartan subalgebra and \(M\) is a
+\(\mathfrak{g}\)-module, does the eigenspace decomposition
\[
- V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda
+ M = \bigoplus_\lambda M_\lambda
\]
-of \(V\) hold? The answer to this question turns out to be yes. This is a
+of \(M\) hold? The answer to this question turns out to be yes. This is a
consequence of something known as \emph{simultaneous diagonalization}, which is
the primary tool we will use to generalize the results of the previous section.
What is simultaneous diagonalization all about then?
@@ -127,8 +128,8 @@ What is simultaneous diagonalization all about then?
We should point out that simultaneous diagonalization \emph{only works in the
finite-dimensional setting}. In fact, simultaneous diagonalization is usually
framed as an equivalent statement about diagonalizable \(n \times n\) matrices.
-Simultaneous diagonalization implies that to show \(V = \bigoplus_\lambda
-V_\lambda\) it suffices to show that \(H\!\restriction_V : V \to V\) is a
+Simultaneous diagonalization implies that to show \(M = \bigoplus_\lambda
+M_\lambda\) it suffices to show that \(H\!\restriction_M : M \to M\) is a
diagonalizable operator for each \(H \in \mathfrak{h}\). To that end, we
introduce \emph{the Jordan decomposition of an operator} and \emph{the abstract
Jordan decomposition of a semisimple Lie algebra}.
@@ -159,16 +160,16 @@ It should be clear from the uniqueness of
\(\operatorname{ad}(X)\) is \(\operatorname{ad}(X) =
\operatorname{ad}(X_{\operatorname{s}}) +
\operatorname{ad}(X_{\operatorname{n}})\). What is perhaps more remarkable is
-the fact this holds for \emph{any} finite-dimensional representation of
-\(\mathfrak{g}\). In other words\dots
+the fact this holds for \emph{any} finite-dimensional \(\mathfrak{g}\)-module.
+In other words\dots
\begin{proposition}\label{thm:preservation-jordan-form}
- Let \(V\) be a finite-dimensional representation of \(\mathfrak{g}\) and \(X
- \in \mathfrak{g}\). Denote by \(X\!\restriction_V\) the action of \(X\) on
- \(V\). Then \(X_{\operatorname{s}}\!\restriction_V =
- (X\!\restriction_V)_{\operatorname{s}}\) and
- \(X_{\operatorname{n}}\!\restriction_V =
- (X\!\restriction_V)_{\operatorname{n}}\).
+ Let \(M\) be a finite-dimensional \(\mathfrak{g}\)-module and \(X
+ \in \mathfrak{g}\). Denote by \(X\!\restriction_M\) the action of \(X\) on
+ \(M\). Then \(X_{\operatorname{s}}\!\restriction_M =
+ (X\!\restriction_M)_{\operatorname{s}}\) and
+ \(X_{\operatorname{n}}\!\restriction_M =
+ (X\!\restriction_M)_{\operatorname{n}}\).
\end{proposition}
This last result is known as \emph{the preservation of the Jordan form}, and a
@@ -177,24 +178,24 @@ implies\dots
\begin{corollary}\label{thm:finite-dim-is-weight-mod}
Let \(\mathfrak{g}\) be a semisimple Lie algebra, \(\mathfrak{h} \subset
- \mathfrak{g}\) be a Cartan subalgebra and \(V\) be any finite-dimensional
- representation of \(\mathfrak{g}\). Then there is a basis \(\{v_1, \ldots,
- v_n\}\) of \(V\) so that each \(v_i\) is simultaneously an eigenvector of all
+ \mathfrak{g}\) be a Cartan subalgebra and \(M\) be any finite-dimensional
+ \(\mathfrak{g}\)-module. Then there is a basis \(\{m_1, \ldots,
+ m_r\}\) of \(M\) so that each \(m_i\) is simultaneously an eigenvector of all
elements of \(\mathfrak{h}\) -- i.e. each element of \(\mathfrak{h}\) acts as
a diagonal matrix in this basis. In other words, there are linear functionals
\(\lambda_i \in \mathfrak{h}^*\) so that
\(
- H v_i = \lambda_i(H) \cdot v_i
+ H m_i = \lambda_i(H) m_i
\)
for all \(H \in \mathfrak{h}\). In particular,
\[
- V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda
+ M = \bigoplus_{\lambda \in \mathfrak{h}^*} M_\lambda
\]
\end{corollary}
\begin{proof}
- Fix some \(H \in \mathfrak{h}\). It suffices to show that \(H\!\restriction_V
- : V \to V\) is a diagonalizable operator.
+ Fix some \(H \in \mathfrak{h}\). It suffices to show that \(H\!\restriction_M
+ : M \to M\) is a diagonalizable operator.
If we write \(H = H_{\operatorname{s}} + H_{\operatorname{n}}\) for the
abstract Jordan decomposition of \(H\), we know
@@ -207,9 +208,9 @@ implies\dots
\(H_{\operatorname{n}}\) is a central element of \(\mathfrak{g}\). Since
\(\mathfrak{g}\) is semisimple, \(H_{\operatorname{n}} = 0\).
Proposition~\ref{thm:preservation-jordan-form} then implies
- \((H\!\restriction_V)_{\operatorname{n}} =
- (H_{\operatorname{n}})\!\restriction_V = 0\), so \(H\!\restriction_V =
- (H\!\restriction_V)_{\operatorname{s}}\) is a diagonalizable operator.
+ \((H\!\restriction_M)_{\operatorname{n}} =
+ H_{\operatorname{n}}\!\restriction_M = 0\), so \(H\!\restriction_M =
+ (H\!\restriction_M)_{\operatorname{s}}\) is a diagonalizable operator.
\end{proof}
We should point out that this last proof only works for semisimple Lie
@@ -219,21 +220,21 @@ semisimple Lie algebras are centerless. In fact,
Corollary~\ref{thm:finite-dim-is-weight-mod} fails even for reductive Lie
algebras. For a counterexample, consider the algebra \(\mathfrak{g} = K\): the
Cartan subalgebra of \(\mathfrak{g}\) is \(\mathfrak{g}\) itself, and a
-\(\mathfrak{g}\)-module is simply a vector space \(V\) endowed with an operator
-\(V \to V\) -- which corresponds to the action of \(1 \in \mathfrak{g}\) on
-\(V\). In particular, if we choose an operator \(V \to V\) which is \emph{not}
-diagonalizable we find \(V \ne 0 = \bigoplus_{\lambda \in \mathfrak{h}^*}
-V_\lambda\).
+\(\mathfrak{g}\)-module is simply a vector space \(M\) endowed with an operator
+\(M \to M\) -- which corresponds to the action of \(1 \in \mathfrak{g}\) on
+\(M\). In particular, if we choose an operator \(M \to M\) which is \emph{not}
+diagonalizable we find \(M \ne \bigoplus_{\lambda \in \mathfrak{h}^*}
+M_\lambda\).
However, Corollary~\ref{thm:finite-dim-is-weight-mod} does work for reductive
-\(\mathfrak{g}\) if we assume that the representation in question is
-irreducible, since central elements of \(\mathfrak{g}\) act on irreducible
-representations as scalar operators. The hypothesis of finite-dimensionality is
-also of huge importance. In the next chapter we will encounter
-infinite-dimensional \(\mathfrak{g}\)-modules for which the eigenspace
-decomposition \(V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda\) fails.
-As a first consequence of Corollary~\ref{thm:finite-dim-is-weight-mod} we
-show\dots
+\(\mathfrak{g}\) if we assume that the \(\mathfrak{g}\)-module \(M\) in
+question is simple, since central elements of \(\mathfrak{g}\) act on simple
+\(\mathfrak{g}\)-modules as scalar operators. The hypothesis of
+finite-dimensionality is also of huge importance. In the next chapter we will
+encounter infinite-dimensional \(\mathfrak{g}\)-modules for which the
+eigenspace decomposition \(M = \bigoplus_{\lambda \in \mathfrak{h}^*}
+M_\lambda\) fails. As a first consequence of
+Corollary~\ref{thm:finite-dim-is-weight-mod} we show\dots
\begin{corollary}
The restriction of the Killing form \(B\) to \(\mathfrak{h}\) is
@@ -242,9 +243,10 @@ show\dots
\begin{proof}
Consider the root space decomposition \(\mathfrak{g} = \mathfrak{g}_0 \oplus
- \bigoplus_\alpha \mathfrak{g}_\alpha\) of the adjoint representation, where
- \(\alpha\) ranges over all nonzero eigenvalues of the adjoint action of
- \(\mathfrak{h}\). We claim \(\mathfrak{g}_0 = \mathfrak{h}\).
+ \bigoplus_\alpha \mathfrak{g}_\alpha\) of the adjoint
+ \(\mathfrak{g}\)-module, where \(\alpha\) ranges over all nonzero eigenvalues
+ of the adjoint action of \(\mathfrak{h}\). We claim \(\mathfrak{g}_0 =
+ \mathfrak{h}\).
Indeed, since \(\mathfrak{h}\) is Abelian, \(\operatorname{ad}(\mathfrak{h})
\mathfrak{h} = 0\) -- i.e. \(\mathfrak{h} \subset \mathfrak{g}_0\). On the
@@ -287,24 +289,24 @@ hardly ever a non-degenerate form.
We now have most of the necessary tools to reproduce the results of the
previous chapter in a general setting. Let \(\mathfrak{g}\) be a
finite-dimensional semisimple algebra with a Cartan subalgebra \(\mathfrak{h}\)
-and let \(V\) be a finite-dimensional irreducible representation of
-\(\mathfrak{g}\). We will proceed, as we did before, by generalizing the
-results of the previous two sections in order. By now the pattern should be
-starting to become clear, so we will mostly omit technical details and proofs
-analogous to the ones on the previous sections. Further details can be found in
-appendix D of \cite{fulton-harris} and in \cite{humphreys}.
+and let \(M\) be a finite-dimensional simple \(\mathfrak{g}\)-module. We will
+proceed, as we did before, by generalizing the results of the previous two
+sections in order. By now the pattern should be starting to become clear, so we
+will mostly omit technical details and proofs analogous to the ones on the
+previous sections. Further details can be found in appendix D of
+\cite{fulton-harris} and in \cite{humphreys}.
\section{The Geometry of Roots and Weights}
We begin our analysis, as we did for \(\mathfrak{sl}_2(K)\) and
-\(\mathfrak{sl}_3(K)\), by investigating the set of roots of and weights of
+\(\mathfrak{sl}_3(K)\), by investigating the locus of roots of and weights of
\(\mathfrak{g}\). Throughout chapter~\ref{ch:sl3} we have seen that the weights
-of any given finite-dimensional representation of \(\mathfrak{sl}_2(K)\) or
+of any given finite-dimensional module of \(\mathfrak{sl}_2(K)\) or
\(\mathfrak{sl}_3(K)\) can only assume very rigid configurations. For instance,
-we have seen that the roots of \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\)
-are symmetric with respect to the origin. In this chapter we will generalize
-most results from chapter~\ref{ch:sl3} regarding the rigidity of the geometry
-of the set of weights of a given representations.
+we have seen that the roots of \(\mathfrak{sl}_2(K)\) and
+\(\mathfrak{sl}_3(K)\) are symmetric with respect to the origin. In this
+chapter we will generalize most results from chapter~\ref{ch:sl3} regarding the
+rigidity of the geometry of the set of weights of a given module.
As for the aforementioned result on the symmetry of roots, this turns out to be
a general fact, which is a consequence of the non-degeneracy of the restriction
@@ -317,10 +319,9 @@ of the Killing form to the Cartan subalgebra.
\begin{proof}
We will start with the first claim. Let \(\alpha\) and \(\beta\) be two
- roots. Notice
- \([\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \subset \mathfrak{g}_{\alpha +
- \beta}\). Indeed, if \(X \in \mathfrak{g}_\alpha\) and \(Y \in
- \mathfrak{g}_\beta\) then
+ roots. Notice \([\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \subset
+ \mathfrak{g}_{\alpha + \beta}\). Indeed, if \(X \in \mathfrak{g}_\alpha\) and
+ \(Y \in \mathfrak{g}_\beta\) then
\[
[H [X, Y]]
= [X, [H, Y]] - [Y, [H, X]]
@@ -331,12 +332,12 @@ of the Killing form to the Cartan subalgebra.
This implies that if \(\alpha + \beta \ne 0\) then \(\operatorname{ad}(X)
\operatorname{ad}(Y)\) is nilpotent: if \(Z \in \mathfrak{g}_\gamma\) then
\[
- (\operatorname{ad}(X) \operatorname{ad}(Y))^n Z
+ (\operatorname{ad}(X) \operatorname{ad}(Y))^r Z
= [X, [Y, [ \ldots, [X, [Y, Z]]] \ldots ]
- \in \mathfrak{g}_{n \alpha + n \beta + \gamma}
+ \in \mathfrak{g}_{r \alpha + r \beta + \gamma}
= 0
\]
- for \(n\) large enough. In particular, \(B(X, Y) =
+ for \(r\) large enough. In particular, \(B(X, Y) =
\operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\). Now if
\(- \alpha\) is not an eigenvalue we find \(B(X, \mathfrak{g}_\beta) = 0\)
for all roots \(\beta\), which contradicts the non-degeneracy of \(B\).
@@ -355,34 +356,34 @@ Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and
\(\mathfrak{sl}_3(K)\) one can show\dots
\begin{proposition}\label{thm:root-space-dim-1}
- The eigenspaces \(\mathfrak{g}_\alpha\) are all \(1\)-dimensional.
+ The root spaces \(\mathfrak{g}_\alpha\) are all \(1\)-dimensional.
\end{proposition}
The proof of the first statement of
Proposition~\ref{thm:weights-symmetric-span} highlights something interesting:
if we fix some eigenvalue \(\alpha\) of the adjoint action of \(\mathfrak{h}\)
on \(\mathfrak{g}\) and a eigenvector \(X \in \mathfrak{g}_\alpha\), then for
-each \(H \in \mathfrak{h}\) and \(v \in V_\lambda\) we find
+each \(H \in \mathfrak{h}\) and \(m \in M_\lambda\) we find
\[
- H (X v)
- = X (H v) + [H, X] v
- = (\lambda + \alpha)(H) \cdot X v
+ H \cdot (X \cdot m)
+ = X H \cdot m + [H, X] \cdot m
+ = (\lambda + \alpha)(H) X \cdot m
\]
-Thus \(X\) sends \(v\) to \(V_{\lambda + \alpha}\). We have encountered this
+Thus \(X\) sends \(m\) to \(M_{\lambda + \alpha}\). We have encountered this
formula twice in these notes: again, we find \(\mathfrak{g}_\alpha\) \emph{acts
-on \(V\) by translating vectors between eigenspaces}. In particular, if we
+on \(M\) by translating vectors between eigenspaces}. In particular, if we
denote by \(\Delta\) the set of all roots of \(\mathfrak{g}\) then\dots
\begin{theorem}\label{thm:weights-congruent-mod-root}
- The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) are
- all congruent modulo the root lattice \(Q = \mathbb{Z} \Delta\) of \(\mathfrak{g}\).
- In other words, all weights of \(V\) lie in the same \(Q\)-coset
- \(t \in \mfrac{\mathfrak{h}^*}{Q}\).
+ The weights of a finite-dimensional simple \(\mathfrak{g}\)-module \(M\) are
+ all congruent modulo the root lattice \(Q = \mathbb{Z} \Delta\) of
+ \(\mathfrak{g}\). In other words, all weights of \(M\) lie in the same
+ \(Q\)-coset \(\xi \in \mfrac{\mathfrak{h}^*}{Q}\).
\end{theorem}
Again, we may leverage our knowledge of \(\mathfrak{sl}_2(K)\) to obtain
-further restrictions on the geometry of the space of weights of \(V\). Namely,
+further restrictions on the geometry of the locus of weights of \(M\). Namely,
as in the case of \(\mathfrak{sl}_3(K)\) we show\dots
\begin{proposition}\label{thm:distinguished-subalgebra}
@@ -395,7 +396,7 @@ as in the case of \(\mathfrak{sl}_3(K)\) we show\dots
\begin{corollary}\label{thm:distinguished-subalg-rep}
For all weights \(\mu\), the subspace
\[
- \bigoplus_k V_{\mu + k \alpha}
+ \bigoplus_k M_{\mu - k \alpha}
\]
is invariant under the action of the subalgebra \(\mathfrak{s}_\alpha\)
and the weight spaces in this string match the eigenspaces of \(h\).
@@ -404,11 +405,11 @@ as in the case of \(\mathfrak{sl}_3(K)\) we show\dots
The proof of Proposition~\ref{thm:distinguished-subalgebra} is very technical
in nature and we won't include it here, but the idea behind it is simple:
recall that \(\mathfrak{g}_\alpha\) and \(\mathfrak{g}_{- \alpha}\) are both
-\(1\)-dimensional, so that \(\dim [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\)
-is at most 1. We check that \([\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]
-\ne 0\) and that no generator of \([\mathfrak{g}_\alpha, \mathfrak{g}_{-
-\alpha}] \ne 0\) is annihilated by \(\alpha\), so that by adjusting scalars we
-can find \(E_\alpha \in \mathfrak{g}_\alpha\) and \(F_\alpha \in
+\(1\)-dimensional, so that \(\dim [\mathfrak{g}_\alpha, \mathfrak{g}_{-
+\alpha}]\) is at most 1. We check that \([\mathfrak{g}_\alpha, \mathfrak{g}_{-
+\alpha}] \ne 0\) and that no generator of \([\mathfrak{g}_\alpha,
+\mathfrak{g}_{- \alpha}]\) is annihilated by \(\alpha\), so that by adjusting
+scalars we can find \(E_\alpha \in \mathfrak{g}_\alpha\) and \(F_\alpha \in
\mathfrak{g}_{- \alpha}\) such that \(H_\alpha = [E_\alpha, F_\alpha]\)
satisfies
\begin{align*}
@@ -419,9 +420,9 @@ satisfies
The elements \(E_\alpha, F_\alpha \in \mathfrak{g}\) are not uniquely
determined by this condition, but \(H_\alpha\) is. As promised, the second
statement of Corollary~\ref{thm:distinguished-subalg-rep} imposes strong
-restrictions on the weights of \(V\). Namely, if \(\lambda\) is a weight,
-\(\lambda(H_\alpha)\) is an eigenvalue of \(h\) on some representation of
-\(\mathfrak{sl}_2(K)\), so it must be an integer. In other words\dots
+restrictions on the weights of \(M\). Namely, if \(\lambda\) is a weight,
+\(\lambda(H_\alpha)\) is an eigenvalue of \(h\) on some
+\(\mathfrak{sl}_2(K)\)-module, so it must be an integer. In other words\dots
\begin{definition}\label{def:weight-lattice}
The lattice \(P = \{ \lambda \in \mathfrak{h}^* : \lambda(H_\alpha) \in
@@ -431,40 +432,40 @@ restrictions on the weights of \(V\). Namely, if \(\lambda\) is a weight,
\end{definition}
\begin{proposition}\label{thm:weights-fit-in-weight-lattice}
- The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) all
- lie in the weight lattice \(P\).
+ The weights of a finite-dimensional simple \(\mathfrak{g}\)-module \(M\) of
+ all lie in the weight lattice \(P\).
\end{proposition}
Proposition~\ref{thm:weights-fit-in-weight-lattice} is clearly analogous to
Corollary~\ref{thm:sl3-weights-fit-in-weight-lattice}. In fact, the weight
lattice of \(\mathfrak{sl}_3(K)\) -- as in Definition~\ref{def:weight-lattice}
-- is precisely \(\mathbb{Z} \langle \alpha_1, \alpha_2, \alpha_3 \rangle\). To
-proceed further, we would like to take \emph{the highest weight of \(V\)} as in
+proceed further, we would like to take \emph{the highest weight of \(M\)} as in
section~\ref{sec:sl3-reps}, but the meaning of \emph{highest} is again unclear
in this situation. We could simply fix a linear function \(\mathbb{Q} P \to
\mathbb{Q}\) -- as we did in section~\ref{sec:sl3-reps} -- and choose a weight
-\(\lambda\) of \(V\) that maximizes this functional, but at this point it is
+\(\lambda\) of \(M\) that maximizes this functional, but at this point it is
convenient to introduce some additional tools to our arsenal. These tools are
called \emph{basis}.
\begin{definition}\label{def:basis-of-root}
- A subset \(\Sigma = \{\beta_1, \ldots, \beta_k\} \subset \Delta\) of linearly
+ A subset \(\Sigma = \{\beta_1, \ldots, \beta_r\} \subset \Delta\) of linearly
independent roots is called \emph{a basis for \(\Delta\)} if, given \(\alpha
- \in \Delta\), there are \(n_1, \ldots, n_k \in \mathbb{N}\) such that
- \(\alpha = \pm(n_1 \beta_1 + \cdots + n_k \beta_k)\).
+ \in \Delta\), there are \(k_1, \ldots, k_r \in \mathbb{N}\) such that
+ \(\alpha = \pm(k_1 \beta_1 + \cdots + k_r \beta_r)\).
\end{definition}
The interesting thing about basis for \(\Delta\) is that they allow us to
-compare weights of a given representation. At this point the reader should be
-asking himself: how? Definition~\ref{def:basis-of-root} isn't exactly all that
-intuitive. Well, the thing is that any choice of basis induces a partial order
-in \(Q\), where elements are ordered by their \emph{heights}.
+compare weights of a given \(\mathfrak{g}\)-module. At this point the reader
+should be asking himself: how? Definition~\ref{def:basis-of-root} isn't exactly
+all that intuitive. Well, the thing is that any choice of basis induces a
+partial order in \(Q\), where elements are ordered by their \emph{heights}.
\begin{definition}
- Let \(\Sigma = \{\beta_1, \ldots, \beta_k\}\) be a basis for \(\Delta\).
- Given \(\alpha = n_1 \beta_1 + \cdots + n_2 \beta_2 \in Q\) with \(n_1,
- \ldots, n_k \in \mathbb{Z}\), we call the number \(\operatorname{ht}(\alpha)
- = n_1 + \cdots + n_k \in \mathbb{Z}\) \emph{the height of \(\alpha\)}. We say
+ Let \(\Sigma = \{\beta_1, \ldots, \beta_r\}\) be a basis for \(\Delta\).
+ Given \(\alpha = k_1 \beta_1 + \cdots + k_r \beta_r \in Q\) with \(k_1,
+ \ldots, k_r \in \mathbb{Z}\), we call the number \(\operatorname{ht}(\alpha)
+ = k_1 + \cdots + k_r \in \mathbb{Z}\) \emph{the height of \(\alpha\)}. We say
that \(\alpha \preceq \beta\) if \(\operatorname{ht}(\alpha) \le
\operatorname{ht}(\beta)\).
\end{definition}
@@ -491,21 +492,21 @@ It should be obvious that the binary relation \(\preceq\) in \(Q\) is a partial
order. In addition, we may compare the elements of a given \(Q\)-coset
\(\lambda + Q\) by comparing their difference with \(0 \in Q\). In other words,
given \(\lambda \in \mu + Q\), we say \(\lambda \preceq \mu\) if \(\lambda -
-\mu \preceq 0\). In particular, since the weights of \(V\) all lie in a single
+\mu \preceq 0\). In particular, since the weights of \(M\) all lie in a single
\(Q\)-coset, we may compare them in this way. Given a basis \(\Sigma\) for
-\(\Delta\) we may take ``the highest weight of \(V\)'' as a maximal weight
-\(\lambda\) of \(V\). The obvious question then is: can we always find a basis
+\(\Delta\) we may take ``the highest weight of \(M\)'' as a maximal weight
+\(\lambda\) of \(M\). The obvious question then is: can we always find a basis
for \(\Delta\)?
\begin{proposition}
There is a basis \(\Sigma\) for \(\Delta\).
\end{proposition}
-The intuition behind the proof of this Proposition is similar to our original
+The intuition behind the proof of this proposition is similar to our original
idea of fixing a direction in \(\mathfrak{h}^*\) in the case of
\(\mathfrak{sl}_3(K)\). Namely, one can show that \(B(\alpha, \beta) \in
\mathbb{Z}\) for all \(\alpha, \beta \in \Delta\), so that the Killing form
-\(B\) restricts to a nondegenerate \(\mathbb{Q}\)-linear form \(\mathbb{Q}
+\(B\) restricts to a nondegenerate \(\mathbb{Q}\)-bilinear form \(\mathbb{Q}
\Delta \times \mathbb{Q} \Delta \to \mathbb{Q}\). We can then fix a nonzero
vector \(\gamma \in \mathbb{Q} \Delta\) and consider the orthogonal projection
\(f : \mathbb{Q} \Delta \to \mathbb{Q} \gamma \cong \mathbb{Q}\). We say a root
@@ -517,58 +518,59 @@ way.
Fix some basis \(\Sigma\) for \(\Delta\), with corresponding decomposition
\(\Delta^+ \cup \Delta^- = \Delta\). Let \(\lambda\) be a maximal weight of
-\(V\). We call \(\lambda\) \emph{the highest weight of \(V\)}, and we call any
-nonzero \(v \in V_\lambda\) \emph{a highest weight vector}. The strategy then
-is to describe all weight spaces of \(V\) in terms of \(\lambda\) and \(v\), as
+\(M\). We call \(\lambda\) \emph{the highest weight of \(M\)}, and we call any
+nonzero \(m \in M_\lambda\) \emph{a highest weight vector}. The strategy then
+is to describe all weight spaces of \(M\) in terms of \(\lambda\) and \(m\), as
in Theorem~\ref{thm:sl3-irr-weights-class}. Unsurprisingly we do so by
reproducing the proof of the case of \(\mathfrak{sl}_3(K)\).
-First, we note that any highest weight vector \(v \in V_\lambda\) is
+First, we note that any highest weight vector \(m \in M_\lambda\) is
annihilated by all positive root spaces, for if \(\alpha \in \Delta^+\) then
-\(E_\alpha v \in V_{\lambda + \alpha}\) must be zero -- or otherwise we would
-have that \(\lambda + \alpha\) is a weight with \(\lambda \prec \lambda +
+\(E_\alpha \cdot m \in M_{\lambda + \alpha}\) must be zero -- or otherwise we
+would have that \(\lambda + \alpha\) is a weight with \(\lambda \prec \lambda +
\alpha\). In particular,
\[
- \bigoplus_{k \in \mathbb{Z}} V_{\lambda - k \alpha}
- = \bigoplus_{k \in \mathbb{N}} V_{\lambda - k \alpha}
+ \bigoplus_{k \in \mathbb{Z}} M_{\lambda - k \alpha}
+ = \bigoplus_{k \in \mathbb{N}} M_{\lambda - k \alpha}
\]
and \(\lambda(H_\alpha)\) is the right-most eigenvalue of the action of \(h\)
-on the \(\mathfrak{sl}_2(K)\)-module \(\bigoplus_k V_{\lambda - k \alpha}\).
+on the \(\mathfrak{sl}_2(K)\)-module \(\bigoplus_k M_{\lambda - k \alpha}\).
This has a number of important consequences. For instance\dots
\begin{corollary}
If \(\alpha \in \Delta^+\) and \(\sigma_\alpha : \mathfrak{h}^* \to
\mathfrak{h}^*\) is the reflection in the hyperplane perpendicular to
- \(\alpha\) with respect to the Killing form, the weights of \(V\) occurring in
- the line joining \(\lambda\) and \(\sigma_\alpha\) are precisely the \(\mu
+ \(\alpha\) with respect to the Killing form, the weights of \(M\) occurring
+ in the line joining \(\lambda\) and \(\sigma_\alpha\) are precisely the \(\mu
\in P\) lying between \(\lambda\) and \(\sigma_\alpha(\lambda)\).
\end{corollary}
\begin{proof}
Notice that any \(\mu \in P\) in the line joining \(\lambda\) and
\(\sigma_\alpha(\lambda)\) has the form \(\mu = \lambda - k \alpha\) for some
- \(k\), so that \(V_\mu\) corresponds the eigenspace associated with the
+ \(k\), so that \(M_\mu\) corresponds the eigenspace associated with the
eigenvalue \(\lambda(H_\alpha) - 2k\) of the action of \(h\) on \(\bigoplus_k
- V_{\lambda - k \alpha}\). If \(\mu\) lies between \(\lambda\) and
+ M_{\lambda - k \alpha}\). If \(\mu\) lies between \(\lambda\) and
\(\sigma_\alpha(\lambda)\) then \(k\) lies between \(0\) and
- \(\lambda(H_\alpha)\), in which case \(V_\mu \neq 0\) and therefore \(\mu\)
+ \(\lambda(H_\alpha)\), in which case \(M_\mu \neq 0\) and therefore \(\mu\)
is a weight.
On the other hand, if \(\mu\) does not lie between \(\lambda\) and
\(\sigma_\alpha(\lambda)\) then either \(k < 0\) or \(k >
\lambda(H_\alpha)\). Suppose \(\mu\) is a weight. In the first case \(\mu
- \succ \lambda\), a contradiction. On the second case the fact that \(V_\mu
- \ne 0\) implies \(V_{\lambda + (k - \lambda(H_\alpha)) \alpha} =
- V_{\sigma_\alpha(\mu)} \ne 0\), which contradicts the fact that \(V_{\lambda
+ \succ \lambda\), a contradiction. On the second case the fact that \(M_\mu
+ \ne 0\) implies \(M_{\lambda + (k - \lambda(H_\alpha)) \alpha} =
+ M_{\sigma_\alpha(\mu)} \ne 0\), which contradicts the fact that \(M_{\lambda
+ \ell \alpha} = 0\) for all \(\ell \ge 0\).
\end{proof}
This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we
-found that the weights of the irreducible representations formed continuous
-strings symmetric with respect to the lines \(K \alpha\) with \(B(\alpha_i -
-\alpha_j, \alpha) = 0\). As in the case of \(\mathfrak{sl}_3(K)\), the same
-class of arguments leads us to the conclusion\dots
+found that the weights of the simple \(\mathfrak{sl}_3(K)\)-modules formed
+continuous strings symmetric with respect to the lines \(K \alpha\) with
+\(B(\alpha_i - \alpha_j, \alpha) = 0\). As in the case of
+\(\mathfrak{sl}_3(K)\), the same class of arguments leads us to the
+conclusion\dots
\begin{definition}
We refer to the group \(\mathcal{W} = \langle \sigma_\alpha : \alpha \in
@@ -577,11 +579,10 @@ class of arguments leads us to the conclusion\dots
\end{definition}
\begin{theorem}\label{thm:irr-weight-class}
- The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) with
- highest weight \(\lambda\) are precisely the elements of the weight lattice
- \(P\) congruent to \(\lambda\) modulo the root lattice \(Q\) lying inside the
- convex hull of the image of \(\lambda\) under the action of the Weyl group
- \(\mathcal{W}\).
+ The weights of a simple \(\mathfrak{g}\)-module \(M\) with highest weight
+ \(\lambda\) are precisely the elements of the weight lattice \(P\) congruent
+ to \(\lambda\) modulo the root lattice \(Q\) lying inside the convex hull of
+ the orbit of \(\lambda\) under the action of the Weyl group \(\mathcal{W}\).
\end{theorem}
Aside from showing up in the previous theorem, the Weyl group will also play an
@@ -628,7 +629,7 @@ roots and weights are only the beginning of a well develop axiomatic theory of
the so called \emph{root systems}, which was used by Cartan in the early 20th
century to classify all finite-dimensional simple complex Lie algebras in terms
of Dynking diagrams. This and much more can be found in \cite[III]{humphreys}
-and \cite[ch.~21]{fulton-harris}. Having found all of the weights of \(V\), the
+and \cite[ch.~21]{fulton-harris}. Having found all of the weights of \(M\), the
only thing we are missing for a complete classification is an existence and
uniqueness theorem analogous to Theorem~\ref{thm:sl2-exist-unique} and
Theorem~\ref{thm:sl3-existence-uniqueness}. This will be the focus of the next
@@ -637,11 +638,11 @@ section.
\section{Verma Modules \& the Highest Weight Theorem}
It is already clear from the previous discussion that if \(\lambda\) is the
-highest weight of \(V\) then \(\lambda(H_\alpha) \ge 0\) for all positive roots
+highest weight of \(M\) then \(\lambda(H_\alpha) \ge 0\) for all positive roots
\(\alpha\). In other words, having \(\lambda(H_\alpha) \ge 0\), for all
-\(\alpha \in \Delta^+\), is a necessary condition for the existence of
-irreducible representations with highest weight given by \(\lambda\).
-Surprisingly, this condition is also sufficient. In other words\dots
+\(\alpha \in \Delta^+\), is a necessary condition for the existence of a simple
+\(\mathfrak{g}\)-module with highest weight given by \(\lambda\). Surprisingly,
+this condition is also sufficient. In other words\dots
\begin{definition}
An element \(\lambda\) of \(P\) such that \(\lambda(H_\alpha) \ge 0\) for all
@@ -651,8 +652,8 @@ Surprisingly, this condition is also sufficient. In other words\dots
\begin{theorem}\label{thm:dominant-weight-theo}
For each dominant integral \(\lambda \in P\) there exists precisely one
- irreducible finite-dimensional representation \(V\) of \(\mathfrak{g}\) whose
- highest weight is \(\lambda\).
+ finite-dimensional simple \(\mathfrak{g}\)-module \(M\) whose highest weight
+ is \(\lambda\).
\end{theorem}
This is known as \emph{the Highest Weight Theorem}, and its proof is the focus
@@ -660,10 +661,10 @@ of this section. The ``uniqueness'' part of the theorem follows at once from
the argument used for \(\mathfrak{sl}_3(K)\). Namely\dots
\begin{proposition}\label{thm:irr-subrep-generated-by-vec}
- Let \(V\) be a finite-dimensional representation of \(\mathfrak{g}\) and \(v
- \in V\) be a highest weight vector. Then the subrepresentation
- \(\mathcal{U}(\mathfrak{g}) \cdot v \subset V\) generated by \(v\) is
- irreducible.
+ Let \(M\) be a finite-dimensional \(\mathfrak{g}\)-module and \(m
+ \in M\) be a highest weight vector. Then the \(\mathfrak{g}\)-submodule
+ \(\mathcal{U}(\mathfrak{g}) \cdot m \subset M\) generated by \(m\) is
+ simple.
\end{proposition}
The proof of Proposition~\ref{thm:irr-subrep-generated-by-vec} is very similar
@@ -675,19 +676,19 @@ applications of negative root vectors is invariant under the action of
Of course, what we are really interested in is\dots
\begin{corollary}
- Let \(V\) and \(W\) be finite-dimensional irreducible
- \(\mathfrak{g}\)-modules with highest weight given by some common \(\lambda
- \in P\). Then \(V \cong W\).
+ Let \(M\) and \(N\) be finite-dimensional simple \(\mathfrak{g}\)-modules
+ with highest weight given by some common \(\lambda \in P\). Then \(M \cong
+ N\).
\end{corollary}
\begin{proof}
- Let \(v \in V\) and \(w \in W\) be highest weight vectors and \(U =
- \mathcal{U}(\mathfrak{g}) \cdot v + w \subset V \oplus W\). It is clear that
- \(v + w\) is a highest weight vector of \(V \oplus W\). Hence by
- Proposition~\ref{thm:irr-subrep-generated-by-vec} \(U\) is irreducible. The
- projections \(\pi_1 : U \to V\) and \(\pi_2 : U \to W\) are thus nonzero
- intertwiners between irreducible representations of \(\mathfrak{g}\) and are
- therefore isomorphisms. Hence \(V \cong U \cong W\).
+ Let \(m \in M\) and \(n \in N\) be highest weight vectors and \(L =
+ \mathcal{U}(\mathfrak{g}) \cdot m + n \subset M \oplus N\). It is clear that
+ \(m + n\) is a highest weight vector of \(M \oplus N\). Hence by
+ Proposition~\ref{thm:irr-subrep-generated-by-vec} \(L\) is simple. The
+ projections \(\pi_1 : L \to M\) and \(\pi_2 : L \to N\) are thus nonzero
+ \(\mathfrak{g}\)-homomorphisms between simple \(\mathfrak{g}\)-modules and
+ are therefore isomorphisms. Hence \(M \cong L \cong N\).
\end{proof}
The ``existence'' part is more nuanced. Our first instinct is, of course, to
@@ -699,20 +700,20 @@ known as \emph{Verma modules}.
\begin{definition}\label{def:verma}
The \(\mathfrak{g}\)-module \(M(\lambda) =
- \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K v^+\), where the action of
- \(\mathfrak{b}\) on \(K v^+\) is given by \(H v^+ = \lambda(H) \cdot v^+\)
- for all \(H \in \mathfrak{h}\) and \(X v^+ = 0\) for \(X \in
+ \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K m^+\), where the action of
+ \(\mathfrak{b}\) on \(K m^+\) is given by \(H \cdot m^+ = \lambda(H) m^+\)
+ for all \(H \in \mathfrak{h}\) and \(X \cdot v^+ = 0\) for \(X \in
\mathfrak{g}_{\alpha}\), \(\alpha \in \Delta^+\), is called \emph{the Verma
- module of weight \(\lambda\)}
+ module of weight \(\lambda\)}.
\end{definition}
-We should point out that, unlike most representations we have encountered so far,
+We should point out that, unlike most modules we have encountered so far,
Verma modules are \emph{highly infinite-dimensional}. Indeed, the dimension of
\(M(\lambda)\) is the same as the codimension of \(\mathcal{U}(\mathfrak{b})\)
in \(\mathcal{U}(\mathfrak{g})\), which is always infinite. Nevertheless,
\(M(\lambda)\) turns out to be quite well behaved. For instance, by
-construction \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\) -- where
-\(v^+ = 1 \otimes v^+ \in M(\lambda)\) is as in Definition~\ref{def:verma}.
+construction \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot m^+\) -- where
+\(m^+ = 1 \otimes m^+ \in M(\lambda)\) is as in Definition~\ref{def:verma}.
Moreover, we find\dots
\begin{proposition}\label{thm:verma-is-weight-mod}
@@ -721,32 +722,33 @@ Moreover, we find\dots
M(\lambda) = \bigoplus_{\mu \in \mathfrak{h}^*} M(\lambda)_\mu
\]
holds. Furthermore, \(\dim M(\lambda)_\mu < \infty\) for all \(\mu \in
- \mathfrak{h}^*\) and \(\dim M(\lambda) = 1\). Finally, \(\lambda\) is the
- highest weight of \(M(\lambda)\), with highest weight vector given by \(v^+ =
- 1 \otimes v^+ \in M(\lambda)\) as in Definition~\ref{def:verma}.
+ \mathfrak{h}^*\) and \(\dim M(\lambda)_\lambda = 1\). Finally, \(\lambda\) is
+ the highest weight of \(M(\lambda)\), with highest weight vector given by
+ \(m^+ = 1 \otimes m^+ \in M(\lambda)\) as in Definition~\ref{def:verma}.
\end{proposition}
\begin{proof}
The Poincaré-Birkhoff-Witt Theorem implies that \(M(\lambda)\) is spanned by
- the vectors \(F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v^+\) for
- \(\alpha_i \in \Delta^-\) and \(F_{\alpha_i} \in \mathfrak{g}_{\alpha_i}\) as
- in the proof of Proposition~\ref{thm:distinguished-subalgebra}. But
+ the vectors \(F_{\alpha_{i_1}} F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot
+ m^+\) for \(\Delta^+ = \{\alpha_1, \ldots, \alpha_r\}\) and \(F_{\alpha_i}
+ \in \mathfrak{g}_{- \alpha_i}\) as in the proof of
+ Proposition~\ref{thm:distinguished-subalgebra}. But
\[
\begin{split}
- H F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v^+
- & = ([H, F_{\alpha_1}] + F_{\alpha_1} H)
- F_{\alpha_2} \cdots F_{\alpha_n} v^+ \\
- & = \alpha_1(H) \cdot F_{\alpha_1} \cdots F_{\alpha_n} v^+
- + F_{\alpha_1} ([H, F_{\alpha_2}] + F_{\alpha_2} H)
- F_{\alpha_2} \cdots F_{\alpha_n} v^+ \\
+ H \cdot (F_{\alpha_{i_1}} F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+)
+ & = ([H, F_{\alpha_{i_1}}] + F_{\alpha_{i_1}} H)
+ F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\
+ & = - \alpha_{i_1}(H) F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+
+ + F_{\alpha_{i_1}} ([H, F_{\alpha_{i_2}}] + F_{\alpha_{i_2}} H)
+ F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\
& \;\; \vdots \\
- & = (\alpha_1 + \cdots + \alpha_n)(H) \cdot
- F_{\alpha_1} \cdots F_{\alpha_n} v^+
- + F_{\alpha_1} \cdots F_{\alpha_n} H v^+ \\
- & = (\lambda + \alpha_1 + \cdots + \alpha_n)(H) \cdot
- F_{\alpha_1} \cdots F_{\alpha_n} v^+ \\
- & \therefore F_{\alpha_1} \cdots F_{\alpha_n} v^+
- \in M(\lambda)_{\lambda + \alpha_1 + \cdots + \alpha_n}
+ & = (- \alpha_{i_1} - \cdots - \alpha_{i_s})(H)
+ F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+
+ + F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} H \cdot m^+ \\
+ & = (\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s})(H)
+ F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\
+ & \therefore F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+
+ \in M(\lambda)_{\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s}}
\end{split}
\]
@@ -756,50 +758,50 @@ Moreover, we find\dots
M(\lambda)
\subset
\bigoplus_{k_i \in \mathbb{N}}
- M(\lambda)_{\lambda + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n}
+ M(\lambda)_{\lambda - k_1 \cdot \alpha_1 - \cdots - k_r \cdot \alpha_r}
\]
- where \(\{\alpha_1, \ldots, \alpha_m\} = \Delta^-\), so that all weights of
- \(M(\lambda)\) have the form \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots +
- k_n \cdot \alpha_n\).
+ where \(\{\alpha_1, \ldots, \alpha_r\} = \Delta^+\), so that all weights of
+ \(M(\lambda)\) have the form \(\mu = \lambda - k_1 \cdot \alpha_1 - \cdots -
+ k_r \cdot \alpha_r\).
This already gives us that the weights of \(M(\lambda)\) are bounded by
\(\lambda\). To see that \(\lambda\) is indeed a weight, we show that
- \(v^+\) is nonzero weight vector. Clearly \(v^+ \in V_\lambda\). The
+ \(m^+\) is nonzero weight vector. Clearly \(m^+ \in M_\lambda\). The
Poincaré-Birkhoff-Witt Theorem implies \(\mathcal{U}(\mathfrak{g})\) is a
- free \(\mathcal{U}(\mathfrak{b})\)-module, so that
+ free \(\mathfrak{b}\)-module, so that
\[
M(\lambda)
\cong \left(\bigoplus_i \mathcal{U}(\mathfrak{b}) \right)
- \otimes_{\mathcal{U}(\mathfrak{b})} K v^+
+ \otimes_{\mathcal{U}(\mathfrak{b})} K m^+
\cong \bigoplus_i \mathcal{U}(\mathfrak{b})
- \otimes_{\mathcal{U}(\mathfrak{b})} K v^+
- \cong \bigoplus_i K v^+
+ \otimes_{\mathcal{U}(\mathfrak{b})} K m^+
+ \cong \bigoplus_i K m^+
\ne 0
\]
- as \(\mathcal{U}(\mathfrak{b})\)-modules. We then conclude \(v^+ \ne 0\) in
+ as \(\mathcal{U}(\mathfrak{b})\)-modules. We then conclude \(m^+ \ne 0\) in
\(M(\lambda)\), for if this was not the case we would find \(M(\lambda) =
- \mathcal{U}(\mathfrak{g}) \cdot v^+ = 0\). Hence \(V_\lambda \ne 0\) and
+ \mathcal{U}(\mathfrak{g}) \cdot m^+ = 0\). Hence \(M_\lambda \ne 0\) and
therefore \(\lambda\) is the highest weight of \(M(\lambda)\), with highest
- weight vector \(v^+\).
+ weight vector \(m^+\).
To see that \(\dim M(\lambda)_\mu < \infty\), simply note that there are only
finitely many monomials \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots
- F_{\alpha_n}^{k_n}\) such that \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots
- + k_n \cdot \alpha_n\). Since \(M(\lambda)_\mu\) is spanned by the images of
- \(v^+\) under such monomials, we conclude \(\dim M(\lambda) < \infty\). In
+ F_{\alpha_s}^{k_s}\) such that \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots
+ + k_s \cdot \alpha_s\). Since \(M(\lambda)_\mu\) is spanned by the images of
+ \(m^+\) under such monomials, we conclude \(\dim M(\lambda) < \infty\). In
particular, there is a single monomials \(F_{\alpha_1}^{k_1}
- F_{\alpha_2}^{k_2} \cdots F_{\alpha_n}^{k_n}\) such that \(\lambda = \lambda
- + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n\) -- which is, of course,
- the monomial where \(k_1 = \cdots = k_n = 0\). Hence \(\dim V_\lambda = 1\).
+ F_{\alpha_2}^{k_2} \cdots F_{\alpha_s}^{k_s}\) such that \(\lambda = \lambda
+ + k_1 \cdot \alpha_1 + \cdots + k_s \cdot \alpha_s\) -- which is, of course,
+ the monomial where \(k_1 = \cdots = k_n = 0\). Hence \(\dim M_\lambda = 1\).
\end{proof}
\begin{example}\label{ex:sl2-verma}
If \(\mathfrak{g} = \mathfrak{sl}_2(K)\), then we can take \(\mathfrak{h} = K
h\) and \(\mathfrak{b} = K e \oplus K h\). If \(\lambda \in
\mathfrak{h}^*\) is the map \(h \mapsto 2\) then \(M(\lambda) =
- \bigoplus_{k \ge 0} K f^k v^+\), and the action of \(\mathfrak{sl}_2(K)\) on
- \(M(\lambda)\) is given by the formulas in (\ref{eq:sl2-verma-formulas}).
- Visually,
+ \bigoplus_{k \ge 0} K f^k \cdot m^+\), and the action of
+ \(\mathfrak{sl}_2(K)\) on \(M(\lambda)\) is given by the formulas in
+ (\ref{eq:sl2-verma-formulas}). Visually,
\begin{center}
\begin{tikzcd}
\cdots \arrow[bend left=60]{r}{-10}
@@ -810,16 +812,16 @@ Moreover, we find\dots
& M(\lambda)_2 \arrow[bend left=60]{l}{1}
\end{tikzcd}
\end{center}
- where \(M(\lambda)_{2 - 2 k} = K f^k v\). Here the top arrows represent the
- action of \(e\) and the bottom arrows represent the action of \(f\). The
- scalars labeling each arrow indicate to which multiple of \(f^{k \pm 1} v\)
- the elements \(e\) and \(f\) send \(f^k v\). The string of weight spaces to
- the left of the diagram is infinite.
+ where \(M(\lambda)_{2 - 2 k} = K f^k \cdot m^+\). Here the top arrows
+ represent the action of \(e\) and the bottom arrows represent the action of
+ \(f\). The scalars labeling each arrow indicate to which multiple of \(f^{k
+ \pm 1} \cdot m^+\) the elements \(e\) and \(f\) send \(f^k \cdot m^+\). The
+ string of weight spaces to the left of the diagram is infinite.
\begin{equation}\label{eq:sl2-verma-formulas}
\begin{aligned}
- f^k v^+ & \overset{e}{\mapsto} (2 - k (k + 1)) f^{k - 1} v^+ &
- f^k v^+ & \overset{f}{\mapsto} f^{k + 1} v^+ &
- f^k v^+ & \overset{h}{\mapsto} (2 - 2k) f^k v^+ &
+ f^k \cdot m^+ & \overset{e}{\mapsto} (2 - k(k + 1)) f^{k - 1} \cdot m^+ &
+ f^k \cdot m^+ & \overset{f}{\mapsto} f^{k + 1} \cdot m^+ &
+ f^k \cdot m^+ & \overset{h}{\mapsto} (2 - 2k) f^k \cdot m^+ &
\end{aligned}
\end{equation}
\end{example}
@@ -827,132 +829,131 @@ Moreover, we find\dots
What is interesting to us about all this is that we have just constructed a
\(\mathfrak{g}\)-module whose highest weight is \(\lambda\). This is not a
proof of Theorem~\ref{thm:dominant-weight-theo}, however, since \(M(\lambda)\)
-is neither irreducible nor finite-dimensional. Nevertheless, we can use
-\(M(\lambda)\) to construct an irreducible representation of \(\mathfrak{g}\)
-whose highest weight is \(\lambda\).
+is neither simple nor finite-dimensional. Nevertheless, we can use
+\(M(\lambda)\) to construct a simple \(\mathfrak{g}\)-module whose highest
+weight is \(\lambda\).
\begin{proposition}\label{thm:max-verma-submod-is-weight}
- Every subrepresentation \(V \subset M(\lambda)\) is the direct sum of its
- weight spaces. In particular, \(M(\lambda)\) has a unique maximal
- subrepresentation \(N(\lambda)\) and a unique irreducible quotient
- \(L(\lambda) = \sfrac{M(\lambda)}{N(\lambda)}\).
+ Every submodule \(N \subset M(\lambda)\) is the direct sum of its weight
+ spaces. In particular, \(M(\lambda)\) has a unique maximal submodule
+ \(N(\lambda)\) and a unique simple quotient \(L(\lambda) =
+ \sfrac{M(\lambda)}{N(\lambda)}\).
\end{proposition}
\begin{proof}
- Let \(V \subset M(\lambda)\) be a subrepresentation and take any nonzero \(v
- \in V\). Because of Proposition~\ref{thm:verma-is-weight-mod}, we know there
- are \(\mu_1, \ldots, \mu_n \in \mathfrak{h}^*\) and nonzero \(v_i \in
- M(\lambda)_{\mu_i}\) such that \(v = v_1 + \cdots + v_n\). We want to show
- \(v_i \in V\) for all \(i\).
+ Let \(N \subset M(\lambda)\) be a submodule and take any nonzero \(n \in N\).
+ Because of Proposition~\ref{thm:verma-is-weight-mod}, we know there are
+ \(\mu_1, \ldots, \mu_r \in \mathfrak{h}^*\) and nonzero \(m_i \in
+ M(\lambda)_{\mu_i}\) such that \(n = m_1 + \cdots + m_r\). We want to show
+ \(m_i \in N\) for all \(i\).
Fix some \(H_2 \in \mathfrak{h}\) such that \(\mu_1(H_2) \ne \mu_2(H_2)\).
Then
\[
- v_1
- - \frac{(\mu_3 - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_3
+ m_1
+ - \frac{(\mu_3 - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} \cdot m_3
- \cdots
- - \frac{(\mu_n - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_n
- = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) v
- \in V
+ - \frac{(\mu_r - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} \cdot m_r
+ = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) \cdot n
+ \in N
\]
Now take \(H_3 \in \mathfrak{h}\) such that \(\mu_1(H_3) \ne \mu_3(H_3)\). By
applying the same procedure again we get
\begin{multline*}
- v_1
+ m_1
-
\frac{(\mu_4 - \mu_3)(H_3) \cdot (\mu_4 - \mu_1)(H_2)}
- {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} v_4
+ {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} \cdot m_4
- \cdots -
- \frac{(\mu_n - \mu_3)(H_3) \cdot (\mu_n - \mu_1)(H_2)}
- {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} v_n \\
+ \frac{(\mu_r - \mu_3)(H_3) \cdot (\mu_r - \mu_1)(H_2)}
+ {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} \cdot m_r \\
=
\left(1 - \frac{H_3 - \mu_1(H_3)}{(\mu_3 - \mu_1)(H_3)} \right)
- \left(1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) v
- \in V
+ \left(1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) \cdot n
+ \in N
\end{multline*}
- By applying the same procedure over and over again we can see that \(v_1 = u
- v \in V\) for some \(u \in \mathcal{U}(\mathfrak{g})\). Furthermore, if we
- reproduce all this for \(v_2 + \cdots + v_n = v - v_1 \in V\) we get that
- \(v_2 \in V\). Now by interating this procedure we find \(v_1, \ldots, v_n
- \in V\). Hence
+ By applying the same procedure over and over again we can see that \(m_1 = u
+ \cdot n \in N\) for some \(u \in \mathcal{U}(\mathfrak{g})\). Furthermore, if
+ we reproduce all this for \(m_2 + \cdots + m_r = n - m_1 \in N\) we get that
+ \(m_2 \in N\). All in all we find \(m_1, \ldots, m_r \in N\). Hence
\[
- V = \bigoplus_\mu V_\mu = \bigoplus_\mu M(\lambda)_\mu \cap V
+ N = \bigoplus_\mu N_\mu = \bigoplus_\mu M(\lambda)_\mu \cap N
\]
- Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\), if \(V\) is a
- proper subrepresentation then \(v^+ \notin V\). Hence any proper submodule
- lies in the sum of weight spaces other than \(M(\lambda)_\lambda\), so the
- sum \(N(\lambda)\) of all such submodules is still proper. This implies
- \(N(\lambda)\) is the unique maximal subrepresentation of \(M(\lambda)\) and
- \(L(\lambda) = \sfrac{M(\lambda)}{N(\lambda)}\) is its unique irreducible
+ Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot m^+\), if \(N\) is a
+ proper submodule then \(m^+ \notin N\). Hence any proper submodule lies in
+ the sum of weight spaces other than \(M(\lambda)_\lambda\), so the sum
+ \(N(\lambda)\) of all such submodules is still proper. This implies
+ \(N(\lambda)\) is the unique maximal submodule of \(M(\lambda)\) and
+ \(L(\lambda) = \sfrac{M(\lambda)}{N(\lambda)}\) is its unique simple
quotient.
\end{proof}
\begin{example}\label{ex:sl2-verma-quotient}
If \(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto 2\), we
can see from Example~\ref{ex:sl2-verma} that \(N(\lambda) = \bigoplus_{k \ge
- 3} K f^k v^+\), so that \(L(\lambda)\) is the \(3\)-dimensional irreducible
- representation of \(\mathfrak{sl}_2(K)\) -- i.e. the finite-dimensional
- irreducible representation with highest weight \(\lambda\) constructed in
- chapter~\ref{ch:sl3}.
+ 3} K f^k \cdot m^+\), so that \(L(\lambda)\) is the \(3\)-dimensional simple
+ \(\mathfrak{sl}_2(K)\)-module -- i.e. the finite-dimensional simple module
+ with highest weight \(\lambda\) constructed in chapter~\ref{ch:sl3}.
\end{example}
-This last example is particularly interesting to us, since it indicates that
-the finite-dimensional irreducible representations of \(\mathfrak{sl}_2(K)\) as
-quotients of Verma modules. This is because the quotient
-\(\sfrac{M(\lambda)}{N(\lambda)}\) in Example~\ref{ex:sl2-verma-quotient}
-is finite-dimensional. As it turns out, this is not a coincidence.
+This last example is particularly interesting to us, since it show we can
+realize the finite-dimensional simple \(\mathfrak{sl}_2(K)\)-module as
+quotients of Verma modules. This is because the quotient \(L(\lambda)\) in
+Example~\ref{ex:sl2-verma-quotient} is finite-dimensional. As it turns out,
+this is not a coincidence.
\begin{proposition}\label{thm:verma-is-finite-dim}
If \(\mathfrak{g}\) is semisimple and \(\lambda\) is dominant integral then
- the unique irreducible quotient of \(M(\lambda)\) is finite-dimensional.
+ the unique simple quotient \(L(\lambda)\) of \(M(\lambda)\) is
+ finite-dimensional.
\end{proposition}
The proof of Proposition~\ref{thm:verma-is-finite-dim} is very technical and we
won't include it here, but the idea behind it is to show that the set of
weights of \(L(\lambda)\) is stable under the natural action of the Weyl group
\(\mathcal{W}\) on \(\mathfrak{h}^*\). One can then show that the every weight
-of \(V\) is conjugate to a single dominant integral weight of
-\(\sfrac{M(\lambda)}{N(\lambda)}\), and that the set of dominant integral
-weights of such irreducible quotient is finite. Since \(W\) is finitely
-generated, this implies the set of weights of the unique irreducible quotient
-of \(M(\lambda)\) is finite. But each weight space is finite-dimensional. Hence
-so is the irreducible quotient.
+of \(L(\lambda)\) is conjugate to a single dominant integral weight of
+\(L(\lambda)\), and that the set of dominant integral weights of \(L(\lambda)\)
+is finite. Since \(\mathcal{W}\) is finitely generated, this implies the set of
+weights of the unique simple quotient of \(M(\lambda)\) is finite. But
+each weight space is finite-dimensional. Hence so is the simple quotient
+\(L(\lambda)\).
We refer the reader to \cite[ch. 21]{humphreys} for further details. What we
are really interested in is\dots
\begin{corollary}
Let \(\lambda\) be a dominant integral weight of \(\mathfrak{g}\). Then there
- is a finite-dimensional irreducible \(\mathfrak{g}\)-module \(V\) whose
- highest weight is \(\lambda\).
+ is a finite-dimensional simple \(\mathfrak{g}\)-module \(M\) whose highest
+ weight is \(\lambda\).
\end{corollary}
\begin{proof}
- Let \(V = L(\lambda)\). It suffices to show that its highest weight is
- \(\lambda\). We have already seen that \(v^+ \in M(\lambda)_\lambda\) is a
- highest weight vector. Now since \(v^+\) lies outside of the maximal
- subrepresentation of \(M(\lambda)\), the projection \(v^+ + N(\lambda) \in
- V\) is nonzero.
+ Let \(M = L(\lambda)\). It suffices to show that its highest weight is
+ \(\lambda\). We have already seen that \(m^+ \in M(\lambda)_\lambda\) is a
+ highest weight vector. Now since \(m^+\) lies outside of the maximal
+ submodule of \(M(\lambda)\), the projection \(m^+ + N(\lambda) \in
+ M\) is nonzero.
- We now claim that \(v^+ + N(\lambda) \in V_\lambda\). Indeed,
+ We now claim that \(m^+ + N(\lambda) \in M_\lambda\). Indeed,
\[
- H (v^+ + N(\lambda))
- = H v^+ + N(\lambda)
- = \lambda(H) \cdot (v^+ + N(\lambda))
+ H \cdot (m^+ + N(\lambda))
+ = H \cdot m^+ + N(\lambda)
+ = \lambda(H) (m^+ + N(\lambda))
\]
- for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of \(V\), with
- weight vector \(v^+ + N(\lambda)\). Finally, we remark that \(\lambda\) is
- the highest weight of \(V\), for if this was not the case we could find a
+ for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of \(M\), with
+ weight vector \(m^+ + N(\lambda)\). Finally, we remark that \(\lambda\) is
+ the highest weight of \(M\), for if this was not the case we could find a
weight \(\mu\) of \(M(\lambda)\) with \(\mu \succ \lambda\).
\end{proof}
We should point out that Proposition~\ref{thm:verma-is-finite-dim} fails for
non-dominant \(\lambda \in P\). While \(\lambda\) is always a maximal weight of
\(M(\lambda)\), one can show that if \(\lambda \in P\) is not dominant then
-\(N(\lambda) = 0\) and \(M(\lambda)\) is irreducible. For instance, if
+\(N(\lambda) = 0\) and \(M(\lambda)\) is simple. For instance, if
\(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto -2\) then the
action of \(\mathfrak{g}\) on \(M(\lambda)\) is given by
\begin{center}
@@ -964,7 +965,7 @@ action of \(\mathfrak{g}\) on \(M(\lambda)\) is given by
& M(\lambda)_{-2} \arrow[bend left=60]{l}{1}
\end{tikzcd},
\end{center}
-so we can see that \(M(\lambda)\) has no proper subrepresentations. Verma
-modules can thus serve as examples of infinite-dimensional irreducible
-representations. Our next question is: what are \emph{all} the
-infinite-dimensional irreducible \(\mathfrak{g}\)-modules?
+so we can see that \(M(\lambda)\) has no proper submodules. Verma modules can
+thus serve as examples of infinite-dimensional simple modules. Our next
+question is: what are \emph{all} the infinite-dimensional simple
+\(\mathfrak{g}\)-modules?